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Finding Maximum and Minimum Values - Problem 3 2,374 views

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

It’s time for one more optimization problem let’s read it. A north-south road intersects east west rail road at point O. A car crosses the tracks heading north at 30 miles per hour a train of 5 miles east of point O is heading west at 40 miles per hour let D be the distance between the train and the car. When is the distance of D at minimum?

Whenever I have problemS that involve intersecting roads these north-south east-west movement I always draw a coordinate system, so let’s start with that. So here is our north south road here is our east west road and let’s put O, point O of the intersection right at the origin.

Now it says that the car crosses the tracks heading north at 30 miles per hour. So I'll put the car somewhere up here let’s say that this a little bit after the car has crossed the tracks. And a train is 5 miles east of point O and it's heading west. And let's say after sometime it's here. We are trying to find when the distance between the car and the train is at minimum. When is that distance in minimum we’ll call this distance D. Let me call the train’s position x, makes sense horizontal axis and the car’s position y.

Now if the car is going 30 miles per hour north, then it makes sense that y is 30 times t, where t would be the time in hours. After it crosses the intersection, t in hours. The train started 5 miles away and it's heading west at 40 miles an hour, so x starts at 5 and then we subtract because it's going to the left 40 t.

So after an hour for example, it will be at -35 way over here. So the distance will be a minimum some time before t equals 0 at the very beginning and when the train crosses. Let’s express the distance between them.

This point has coordinates x, 0 and this point has coordinates 0, y. And so the distance between the two points D is the square root of. And then we need the difference of x coordinates, x minus 0 squared plus the difference of y coordinates. 0 minus y squared and that’s going to be the square root of x² plus y². But of course x and y are both function of t. X² is 5 minus 40t² and y² is 30t². Let’s expand these and see what it simplifies to.

Now 5 minus 40t² is going to be 25 minus, now the product of these two is -200t so twice that -400 t and then the square if this plus 1600t². And then the square of 30t² is 900t².

Let’s simplify that we have 25 minus 400t plus 2500t². That was pretty hard so we have to find when this distance is a minimum. The value of t when this distance is a minimum. In our previous example we use the trick we are about to use now. The idea that when this quadratic is a minimum the square root of it will also be a minimum. So the distance is minimized precisely when this quadratic is a minimum.

So this thing is a min when, and when our quadratics had a minimum. Now this quadratic if you graphed it would be an upward opening quadratic, it’s got the positive leading coefficient it's a minimum at its vertex. So when t equals -b over 2a.

Now the vertex let’s see our b value is -400, our a value you remember a is the coefficient for the square, the t² term so this it’s going to be the 2500. It's negative -400 over 2 times 2500. And that’s 400 over 5000. And of course t is an hour so this is in hours. Well that reduces a lot to 4 over 50 which is about 8 over 100.08 hours. Okay what’s that in minutes?

Well all we have to do is multiply this by 60 and we get .48 minutes, I’m sorry that’s times 6, 4.8. I’ve got to stay on my toes here. 4.8 minutes so not quite 5 minutes after they start. Train is moving to the left the car is moving is moving up, 5 minutes after they start the distance between distance between them is closest.

Now the problem doesn’t ask us what that distance is it just asks when the distance is at minimum so we could stop here. But if we needed to calculate that you would take the .08 hours and plug it back in to your distance function.

That’s it so it's kind of a new trick that whenever you get the square root of a quadratic you can use the fact that the quadratic has a max or a min mean at its vertex.

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